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Questions about the branch of algebra that deals with groups.

5 votes
2 answers
425 views

Centre of group with deficiency at least two (Progress on Murasugi's conjecture)

In 1965, Murasugi [1] conjectured that any finitely presented group with deficiency at least two has trivial centre. The year before, he had proved it true for one-relator groups, and in [1] he proved …
8 votes
1 answer
348 views

Finite two-relator groups and quotients of knot groups

Let $G$ be a one-relator group $\langle A \mid R = 1 \rangle$. Then clearly $G$ is finite if and only if it is cyclic of finite order, i.e. can be given by a presentation $\langle a \mid a^n = 1 \rang …
5 votes

Prove these are not surface groups

Here's an argument that uses only basic combinatorial group theory (Reidemeister-Schreier). Let $n \geq 2$, and let $G = \langle a_1, b_1, \dots, a_g, b_g | [a_1, b_1]^n [a_2, b_2] \cdots [a_g, b_g] \ …
Carl-Fredrik Nyberg Brodda's user avatar
9 votes
3 answers
490 views

Residually solvable Bianchi groups

Let $d$ be a square-free positive integer, and let $\mathcal{O}_d$ be the ring of integers of the quadratic imaginary number field $\mathbb{Q}(\sqrt{-d})$. Consider the Bianchi group $\Gamma_d = \oper …
7 votes

Presentation of special linear group over localizations of the integers

In my recent paper (arXiv:2401.08146), I give a new presentation for $\operatorname{SL}_2(\mathbf{Z}[\frac{1}{2}])$. This group is generated by the two matrices $$ A = \begin{pmatrix}1 & 0 \\ 1 & 1\en …
Carl-Fredrik Nyberg Brodda's user avatar
20 votes
1 answer
977 views

Proof of CFSG assuming every simple group is two-generated

It is well-known that one of the corollaries of the classification of finite simple groups (CFSG) is that every finite simple group can be generated by two elements. In a comment on an answer to an ol …
24 votes

Recognizing free groups

As indicated in the comments, it's undecidable in general to take as input a finite presentation of a group and try to output whether or not this group is free or not. This is a direct consequence of …
Carl-Fredrik Nyberg Brodda's user avatar
10 votes

Analogous results in geometric group theory and Riemannian geometry?

Here is a very classical example. As stated in the comments, Gromov was an early proponent of importing ideas from geometry to group theory, but already thirty years earlier there was work in this dir …
Carl-Fredrik Nyberg Brodda's user avatar
18 votes
1 answer
750 views

Is solvability semi-decidable?

Let $G = \langle A \mid R \rangle$ be a finitely presented group, given by a finite presentation. If $G$ is abelian, then we can verify this fact: simply verify the fact that $[a, b] = 1$ for all gene …
11 votes
Accepted

Cohomology of $\operatorname{GL}_3(\mathbb{F}_2)$

As you mention in your update, you have a general answer, but if you want a concrete answer for the low-dimensional integral cohomology of $G = \operatorname{GL}(3,2)$ (or any other finite group!), yo …
Daniel Asimov's user avatar
9 votes
Accepted

Is the isomorphism problem solvable for torsion-free groups?

Novikov's centrally-symmetric group $\mathfrak{A}_P$ is a torsion-free group with undecidable word problem, constructed in [1]. Novikov did not prove it is torsion-free but, as Adian points out in [Ad …
Carl-Fredrik Nyberg Brodda's user avatar
8 votes
2 answers
480 views

Subgroup membership problem in simple groups

Let $G$ be a finitely presented simple group. By Kuznetsov (1958), $G$ has decidable word problem. However, by Scott [1], $G$ may have undecidable conjugacy problem. Is anything known about other deci …
7 votes

Subgroup membership problem in simple groups

After some digging, I was able to find that the answer to my question exists: the problem can be undecidable. Rattaggi, in an unpublished manuscript (available here), proved that there exists a finite …
Carl-Fredrik Nyberg Brodda's user avatar
3 votes

Which groups are LERF?

Polycyclic groups are LERF, by Mal'cev 1948. In particular, all nilpotent and all abelian groups are LERF. As mentioned in the comments, as not all one-relator groups are residually finite, not all on …
Ashot Minasyan's user avatar
6 votes
Accepted

Is there a residually nilpotent one-relator group that is not residually a finite p-group fo...

The answer is yes, such a group exists: this is the main result of [1]. In fact, one can take the Baumslag-Solitar group $\operatorname{BS}(p^r, -p^r) = \langle a, b \mid ba^{p^r}b^{-1} = a^{-p^r} \ra …
Carl-Fredrik Nyberg Brodda's user avatar

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